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1 Acids, Bases and Salts
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2 Acid and Bases
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5 Acids Review Have a sour taste. Vinegar is a solution of acetic acid. Citrus fruits contain citric acid. React with certain metals to produce hydrogen gas. React with carbonates and bicarbonates to produce carbon dioxide gas Have a bitter taste. Feel slippery. Many soaps contain bases. Bases Review
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6 Some Properties of Acids Review þ Produce H + (as H 3 O + ) ions in water (the hydronium ion is a hydrogen ion attached to a water molecule) þ Taste sour þ Gives a stinging sensation upon touch þ Corrode metals þ Electrolytes þ React with bases to form a salt and water þ pH is less than 7 þ Turns blue litmus paper to red “Blue to Red ACID” þ Phenolphthalein colorless in acidic solutions
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7 Some Properties of Bases Review Produce OH - ions in water Taste bitter, chalky Unreactive with metals Are electrolytes Feel soapy, slippery React with acids to form salts and water pH greater than 7 Turns red litmus paper to blue “Basic Blue”
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8 Acid/Base Definitions Review Definition #1: Arrhenius (traditional) Acids – produce H + ions (or hydronium ions H 3 O + ) Bases – produce OH - ions (problem: some bases don’t have hydroxide ions!- NH 3 )
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9 Arrhenius acid is a substance that produces H + (H 3 O + ) in water Arrhenius base is a substance that produces OH - in water
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10 Acid/Base Definitions Review Definition #2: Brønsted – Lowry Acids – proton donor Bases – proton acceptor A “proton” is really just a hydrogen atom (H + ) that has lost it’s electron!
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11 A Brønsted-Lowry acid is a proton donor A Brønsted-Lowry base is a proton acceptor acid conjugate base base conjugate acid
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12 Acid Nomenclature Review No Oxygen w/Oxygen An easy way to remember which goes with which… “In the cafeteria, you ATE something ICky”
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13 Acid Nomenclature Flowchart
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14 HBr (aq)HBr (aq) H 2 CO 3H 2 CO 3 H 2 SO 3H 2 SO 3 hydrobromic acid carbonic acid sulfurous acid Acid Nomenclature Review
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15 Name ‘Em! HI (aq)HI (aq) HCl (aq)HCl (aq) H 2 SO 3H 2 SO 3 HNO 3HNO 3 HIO 4HIO 4
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16 Indicators An indicator is a chemical that will change ___________ when placed in an acidic, basic or neutral environment. Indicator Colors For Acids litmus paper = _______ phenolphthalein = ___________ red cabbage juice (universal indicator) = ________ methyl orange = _______ Bromthymol blue = ______________ colors red clear Pink red yellow
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17 Indicator Colors for Bases litmus paper = _______ methyl orange = ____________ red cabbage juice (universal indicator) =________ phenolphthalein = ______ Bromthymol blue = _________ AcidBase phenolphthalein blue yellow Green pink blue
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18 How to measure pH with wide-range paper 1. Moisten the pH indicator paper strip with a few drops of solution, by using a stirring rod. 2.Compare the color to the chart on the vial – then read the pH value.
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19 Acids Affect Indicators, by changing their color Blue litmus paper turns red in contact with an acid (and red paper stays red).
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20 Bases Affect Indicators Red litmus paper turns blue in contact with a base (and blue paper stays blue). Phenolphthalein turns purple in a base.
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21 pH Paper : Indicator Colors Acidic Basic Neutral
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22 Some of the many pH Indicators and their pH range
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23 The pH scale is a way of expressing the strength of acids and bases. Instead of using very small numbers, we just use the NEGATIVE power of 10 on the Molarity of the H + (or OH - ) ion. Remember pH means __? Under 7 = acid 7 = neutral Over 7 = base
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24 Strength Acids and Bases are classified according to the degree to which they ionize in water: –Strong are completely ionized (dissociate) in aqueous solution; this means they ionize 100 % –Weak ionize only slightly in aqueous solution Strength is very different from Concentration
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25 Strength vs. Concentration The words concentrated and dilute tell how much of an acid or base is dissolved in solution - refers to the number of moles of acid or base in a given volume The words strong and weak refer to the extent of ionization of an acid or base Is a concentrated, weak acid possible?
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26 Hydrogen Ions from Water Water ionizes, or falls apart into ions: H 2 O ↔ H 1+ + OH 1- Called the “self ionization” of water Occurs to a very small extent: [H 1+ ] = [OH 1- ] = 1 x 10 -7 M Since they are equal, a neutral solution results from water K w = [H 1+ ] x [OH 1- ] = 1 x 10 -14 M 2 K w is called the “ion product constant” for water
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27 Calculating pH and pOH The pH of a solution of an acid or base is directly related to the [H + ] concentration. An acid ionizes to produce [H + ] in solution and a base ionizes to produce [OH - ]. The pOH is directly related to the [OH - ] concentration.
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28 Ion Product Constant H 2 O ↔ H 1+ + OH 1- K w is constant in every aqueous solution: [H + ] x [OH - ] = 1 x 10 -14 M 2 If [H + ] > 10 -7 then [OH - ] < 10 -7 If [H + ] 10 -7 If we know one, other can be determined If [H + ] > 10 -7, it is acidic and [OH - ] < 10 -7 If [H + ] 10 -7 –Basic solutions also called “alkaline”
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29 The pH concept – from 0 to 14 pH = pouvoir hydrogene (Fr.) “hydrogen power” definition: pH = -log[H + ] in neutral pH = -log(1 x 10 -7 ) = 7 in acidic solution [H + ] > 10 -7 pH = -log[H + ] –pH < 7 (from 0 to 7 is the acid range) –in base, pH > 7 (7 to 14 is base range)
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30 Calculating pOH pOH = -log [OH - ] [H + ] x [OH - ] = 1 x 10 -14 M 2 pH + pOH = 14 Thus, a solution with a pOH less than 7 is basic; with a pOH greater than 7 is an acid Not greatly used like pH is.
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31 pH calculations – Solving for H+ If the pH of Coke is 3.12, [H + ] = ??? Because pH = - log [H + ] then - pH = log [H + ] Take antilog (10 x ) of both sides and get 10 -pH = [H + ] [H + ] = 10 -3.12 = 7.6 x 10 -4 M *** to find antilog on your calculator, look for “Shift” or “2 nd function” and then the log button
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32 Example: What is the pH of a solution with a hydrogen ion concentration of 4.2 x 10 -9 M? Key into your calculator [−][log][number] [=] If the number is in scientific notation, key into your calculator [−][log][number][EE][−][exponent] [=] Example from above 4.2 x 10 -9 [−][log][[4.2][EE][-][9] [=] pH = 8.38 To convert[H + ] to pH or [OH - ] to pOH Using A Calculator
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33 Key into your calculator [2 nd ][log][−][number][=] Example from above pH = 6.35 [2 nd ][log][−][6.35][=] What is the hydrogen ion concentration of a solution with a pH of 6.35? [H + ] = 4.47 x 10 -7 To convert pH to [H + ] or pOH to [OH - ] Using A Calculator
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34 Practice Problems: 1) a) Calculate the pH of a 0.001 M HCl solution b) What is the pOH of this solution? c) What is the concentration of [OH − ] ions in the solution? 2) a) Calculate the pOH of a NaOH solution that has a pH of 8.50 b) What is the [OH − ] of this solution? c) What is the concentration of [H + ] ions in the solution? [H + ] = 0.001 MSo…pH = − (log 0.001 M)pH = 3 pH + pOH = 14So…14 − 3 = pOHpOH = 11 [OH − ] = 10 −pOH [OH − ] = 10 −11 Molar or 1 x 10 −11 M So…14 − 8.5 = pOH pH + pOH = 14 [OH − ] = 10 −5.5 Molar or 3.16 x 10 −6 M[OH − ] = 10 −pOH [H + ] = 10 −pH [H + ] = 10 −8.5 Molar or 3.16 x 10 −9 M pOH = 5.5
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35 Let’s Practice!!! Example #1 1)What is the pH and pOH of a 1.2 x 10 -3 M HBr solution? pH: 2.9pOH: 11.1 2)What is the pH and pOH of a 2.34 x 10 -5 M NaOH solution? pOH: 4.6pH: 9.4
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36 More About Water H 2 O can function as both an ACID and a BASE. In pure water there can be AUTOIONIZATION Equilibrium constant for water = K w K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C
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37 More About Water K w = [H 3 O + ] [OH - ] = 1.00 x 10 -14 at 25 o C In a neutral solution [H 3 O + ] = [OH - ] so K w = [H 3 O + ] 2 = [OH - ] 2 and so [H 3 O + ] = [OH - ] = 1.00 x 10 -7 M Autoionization
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38 pOH Since acids and bases are opposites, pH and pOH are opposites!Since acids and bases are opposites, pH and pOH are opposites! pOH does not really exist, but it is useful for changing bases to pH.pOH does not really exist, but it is useful for changing bases to pH. pOH looks at the perspective of a basepOH looks at the perspective of a base pOH = - log [OH - ] Since pH and pOH are on opposite ends, pH + pOH = 14
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39 pH [H + ] [OH - ] pOH
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40 [H 3 O + ], [OH - ] and pH What is the pH of the 0.0010 M NaOH solution? [OH-] = 0.0010 (or 1.0 X 10 -3 M) pOH = - log 0.0010 pOH = - log 0.0010 pOH = 3 pOH = 3 pH = 14 – 3 = 11 OR K w = [H 3 O + ] [OH - ] [HO + ] = 1.0 x 10 -11 M [H 3 O + ] = 1.0 x 10 -11 M pH = - log (1.0 x 10 -11 ) = 11.00
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41 The pH of rainwater collected in a certain region of the northeastern United States on a particular day was 4.82. What is the H + ion concentration of the rainwater? The OH - ion concentration of a blood sample is 2.5 x 10 -7 M. What is the pH of the blood?
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42 [OH - ] [H + ] pOH pH 10 -pOH 10 -pH -Log[H + ] Log[OH - ] -Log[OH - ] 14 - pOH 14 - pH 1.0 x 10 -14 [OH - ] [OH - ] 1.0 x 10 -14 [H + ] [H + ]
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43 Calculating [H + ], pH, [OH - ], and pOH Problem 1: A chemist dilutes concentrated hydrochloric acid to make two solutions: (a) 3.0 M and (b) 0.0024 M. Calculate the [H 3 O + ], pH, [OH - ], and pOH of the two solutions at 25°C. Problem 2: What is the [H + ], [OH - ], and pOH of a solution with pH = 3.67? Is this an acid, base, or neutral? Problem 3: Problem #2 with pH = 8.05?
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44 HNO 3, HCl, H 2 SO 4 and HClO 4 are among the only known strong acids. Strong and Weak Acids/Bases The strength of an acid (or base) is determined by the amount of IONIZATION.
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45 Strong and Weak Acids/Bases Generally divide acids and bases into STRONG or WEAK ones.Generally divide acids and bases into STRONG or WEAK ones. STRONG ACID: HNO 3 (aq) + H 2 O (l) ---> H 3 O + (aq) + NO 3 - (aq) HNO 3 is about 100% dissociated in water.
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46 Weak acids are much less than 100% ionized in water.Weak acids are much less than 100% ionized in water. One of the best known is acetic acid = CH 3 CO 2 H Strong and Weak Acids/Bases
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47 Strong Base: 100% dissociated in water.Strong Base: 100% dissociated in water. NaOH (aq) ---> Na + (aq) + OH - (aq) NaOH (aq) ---> Na + (aq) + OH - (aq) Strong and Weak Acids/Bases Other common strong bases include KOH and Ca(OH) 2. CaO (lime) + H 2 O --> Ca(OH) 2 (slaked lime) Ca(OH) 2 (slaked lime) CaO
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48 Weak base: less than 100% ionized in waterWeak base: less than 100% ionized in water One of the best known weak bases is ammonia NH 3 (aq) + H 2 O (l) NH 4 + (aq) + OH - (aq) Strong and Weak Acids/Bases
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49 Weak Bases
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50 Equilibria Involving Weak Acids and Bases Consider acetic acid, HC 2 H 3 O 2 (HOAc) HC 2 H 3 O 2 + H 2 O H 3 O + + C 2 H 3 O 2 - Acid Conj. base (K is designated K a for ACID) K gives the ratio of ions (split up) to molecules (don’t split up)
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51 Ionization Constants for Acids/Bases Acids ConjugateBases Increase strength
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52 Equilibrium Constants for Weak Acids Weak acid has K a < 1 Leads to small [H 3 O + ] and a pH of 2 - 7
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53 Equilibrium Constants for Weak Bases Weak base has K b < 1 Leads to small [OH - ] and a pH of 12 - 7
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54 Relation of K a, K b, [H 3 O + ] and pH
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55 Equilibria Involving A Weak Acid You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. Step 1. Define equilibrium concs. in ICE table. [HOAc][H 3 O + ][OAc - ] [HOAc][H 3 O + ][OAc - ]initialchangeequilib 1.0000 -x+x+x 1.00-xxx
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56 Equilibria Involving A Weak Acid Step 2. Write K a expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. This is a quadratic. Solve using quadratic formula. or you can make an approximation if x is very small! (Rule of thumb: 10 -5 or smaller is ok)
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57 Equilibria Involving A Weak Acid Step 3. Solve K a expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. First assume x is very small because K a is so small. Now we can more easily solve this approximate expression.
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58 Equilibria Involving A Weak Acid Step 3. Solve K a approximate expression You have 1.00 M HOAc. Calc. the equilibrium concs. of HOAc, H 3 O +, OAc -, and the pH. x = [ H 3 O + ] = [ OAc - ] = 4.2 x 10 -3 M pH = - log [ H 3 O + ] = -log (4.2 x 10 -3 ) = 2.37
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59 Equilibria Involving A Weak Acid Calculate the pH of a 0.0010 M solution of formic acid, HCO 2 H. HCO 2 H + H 2 O HCO 2 - + H 3 O + HCO 2 H + H 2 O HCO 2 - + H 3 O + K a = 1.8 x 10 -4 Approximate solution [H 3 O + ] = 4.2 x 10 -4 M, pH = 3.37 [H 3 O + ] = 4.2 x 10 -4 M, pH = 3.37 Exact Solution [H 3 O + ] = [HCO 2 - ] = 3.4 x 10 -4 M [H 3 O + ] = [HCO 2 - ] = 3.4 x 10 -4 M [HCO 2 H] = 0.0010 - 3.4 x 10 -4 = 0.0007 M [HCO 2 H] = 0.0010 - 3.4 x 10 -4 = 0.0007 M pH = 3.47 pH = 3.47
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60 Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O NH 4 + + OH - NH 3 + H 2 O NH 4 + + OH - K b = 1.8 x 10 -5 Step 1. Define equilibrium concs. in ICE table [NH 3 ][NH 4 + ][OH - ] [NH 3 ][NH 4 + ][OH - ]initialchangeequilib 0.01000 -x+x+x 0.010 - xx x
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61 Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O NH 4 + + OH - NH 3 + H 2 O NH 4 + + OH - K b = 1.8 x 10 -5 Step 1. Define equilibrium concs. in ICE table [NH 3 ][NH 4 + ][OH - ] [NH 3 ][NH 4 + ][OH - ]initialchangeequilib 0.01000 -x+x+x 0.010 - xx x
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62 Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O NH 4 + + OH - NH 3 + H 2 O NH 4 + + OH - K b = 1.8 x 10 -5 Step 2. Solve the equilibrium expression Assume x is small, so x = [OH - ] = [NH 4 + ] = 4.2 x 10 -4 M x = [OH - ] = [NH 4 + ] = 4.2 x 10 -4 M and [NH 3 ] = 0.010 - 4.2 x 10 -4 ≈ 0.010 M The approximation is valid!
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63 Equilibria Involving A Weak Base You have 0.010 M NH 3. Calc. the pH. NH 3 + H 2 O NH 4 + + OH - NH 3 + H 2 O NH 4 + + OH - K b = 1.8 x 10 -5 Step 3. Calculate pH [OH - ] = 4.2 x 10 -4 M so pOH = - log [OH - ] = 3.37 Because pH + pOH = 14, pH = 10.63
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64 Types of Acid/Base Reactions: Summary
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65 pH testing There are several ways to test pHThere are several ways to test pH –Blue litmus paper (red = acid) –Red litmus paper (blue = basic) –pH paper (multi-colored) –pH meter (7 is neutral, 7 base) –Universal indicator (multi-colored) –Indicators like phenolphthalein –Natural indicators like red cabbage, radishes
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66 Paper testing Paper tests like litmus paper and pH paperPaper tests like litmus paper and pH paper –Put a stirring rod into the solution and stir. –Take the stirring rod out, and place a drop of the solution from the end of the stirring rod onto a piece of the paper –Read and record the color change. Note what the color indicates. –You should only use a small portion of the paper. You can use one piece of paper for several tests.
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67 pH paper
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68 pH meter Tests the voltage of the electrolyteTests the voltage of the electrolyte Converts the voltage to pHConverts the voltage to pH Very cheap, accurateVery cheap, accurate Must be calibrated with a buffer solutionMust be calibrated with a buffer solution
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69 pH indicators Indicators are dyes that can be added that will change color in the presence of an acid or base. Some indicators only work in a specific range of pH Once the drops are added, the sample is ruined Some dyes are natural, like radish skin or red cabbage
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70 35.62 mL of NaOH is neutralized with 25.2 mL of 0.0998 M HCl by titration to an equivalence point. What is the concentration of the NaOH? LAB PROBLEM #1: Standardize a solution of NaOH — i.e., accurately determine its concentration.
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71 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Add water to the 3.0 M solution to lower its concentration to 0.50 M Dilute the solution!
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72 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? But how much water do we add?
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73 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do ? How much water is added? The important point is that ---> moles of NaOH in ORIGINAL solution = moles of NaOH in FINAL solution moles of NaOH in FINAL solution
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74 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Amount of NaOH in original solution = M V = M V = (3.0 mol/L)(0.050 L) = 0.15 mol NaOH Amount of NaOH in final solution must also = 0.15 mol NaOH Volume of final solution = (0.15 mol NaOH) / (0.50 M) = 0.30 L or 300 mL
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75 PROBLEM: You have 50.0 mL of 3.0 M NaOH and you want 0.50 M NaOH. What do you do? Conclusion: add 250 mL of water to 50.0 mL of 3.0 M NaOH to make 300 mL of 0.50 M NaOH.
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76 A shortcut A shortcut M 1 V 1 = M 2 V 2 Preparing Solutions by Dilution
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77 You try this dilution problem You have a stock bottle of hydrochloric acid, which is 12.1 M. You need 400 mL of 0.10 M HCl. How much of the acid and how much water will you need?
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